Magnitude of force acting on wedge and block

[dl447342]

Homework Statement

A block of mass m sits on a wedge of mass M, which is free to move along a horizontal surface with negligible friction. The coefficient of static friction between the block and the wedge is $\mu_s$. A horizontal force of a magnitude $F$ pulls on the wedge, as shown in the attached diagram. When $\tan \theta > \mu_s$, which of the following are true:

A) The force F must be directed to the right with a minimum magnitude.

B) The force F must be directed to the right with a maximum magnitude.

C) The block will slip even if $F=0$.

Relevant Equations

Newton's second laws are crucial here. From drawing a free-body diagram, one can deduce that the static friction force $f_s$ satisfies $f_s = m(g\sin \theta +\frac{F}{M+m} \cos\theta) \leq \mu_s m(g\cos\theta - \frac{F}{m+M} \sin\theta)$ and that the required maximum force is $\frac{\mu_s -\tan\theta}{1+\mu_s \tan\theta}(M+m)g$ (I'm stating this because the key part of this question is to determine which of the three statements are correct; I've already figured out the free-body diagrams and Newton's equations).

Clearly if $F = 0$ and $\tan\theta > \mu_s$, then using the above equations for $f_s$ and $n$, we get $f_s > \mu_s n$ so the block will slip. However, it seems that as long as the force $F$ is directed to the right with a certain minimum magnitude, namely $\frac{\tan\theta - \mu_s}{1+\mu_s \tan\theta}(M+m)g$, the block won't slip.

Is this reasoning right?

 

[haruspex]

You have only considered one direction of slip, but I agree with your result.
Are you allowed to choose more than one option? It does say "are true", not "is true".
If it is supposed to be only one, I suspect it was intended to state the inequality the other way around.

 

[Lnewqban]

Those maximum and minimum magnitudes seem ambiguous as a question.